Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $k = \dfrac{q - 10}{-3q + 27} \times \dfrac{q^2 + 14q + 40}{q^2 - 100} $
First factor out any common factors. $k = \dfrac{q - 10}{-3(q - 9)} \times \dfrac{q^2 + 14q + 40}{q^2 - 100} $ Then factor the quadratic expressions. $k = \dfrac {q - 10} {-3(q - 9)} \times \dfrac {(q + 10)(q + 4)} {(q + 10)(q - 10)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {(q - 10) \times (q + 10)(q + 4) } {-3(q - 9) \times (q + 10)(q - 10) } $ $k = \dfrac {(q + 10)(q + 4)(q - 10)} {-3(q + 10)(q - 10)(q - 9)} $ Notice that $(q + 10)$ and $(q - 10)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {\cancel{(q + 10)}(q + 4)(q - 10)} {-3\cancel{(q + 10)}(q - 10)(q - 9)} $ We are dividing by $q + 10$ , so $q + 10 \neq 0$ Therefore, $q \neq -10$ $k = \dfrac {\cancel{(q + 10)}(q + 4)\cancel{(q - 10)}} {-3\cancel{(q + 10)}\cancel{(q - 10)}(q - 9)} $ We are dividing by $q - 10$ , so $q - 10 \neq 0$ Therefore, $q \neq 10$ $k = \dfrac {q + 4} {-3(q - 9)} $ $ k = \dfrac{-(q + 4)}{3(q - 9)}; q \neq -10; q \neq 10 $